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Using congruent triangles, one can prove that the rhombus is symmetric across each of these diagonals. It follows that any rhombus has the following properties: Opposite angles of a rhombus have equal measure. The two diagonals of a rhombus are perpendicular; that is, a rhombus is an orthodiagonal quadrilateral. Its diagonals bisect opposite ...
By using the area formula of the general rhombus in terms of its diagonal lengths and : The area of the golden rhombus in terms of its diagonal length d {\displaystyle d} is: [ 6 ] A = ( φ d ) ⋅ d 2 = φ 2 d 2 = 1 + 5 4 d 2 ≈ 0.80902 d 2 . {\displaystyle A={{(\varphi d)\cdot d} \over 2}={{\varphi } \over 2}~d^{2}={{1+{\sqrt {5}}} \over 4 ...
Bretschneider's formula generalizes Brahmagupta's formula for the area of a cyclic quadrilateral, which in turn generalizes Heron's formula for the area of a triangle.. The trigonometric adjustment in Bretschneider's formula for non-cyclicality of the quadrilateral can be rewritten non-trigonometrically in terms of the sides and the diagonals e and f to give [2] [3]
The area K of an orthodiagonal quadrilateral equals one half the product of the lengths of the diagonals p and q: [7] K = p q 2 . {\displaystyle K={\frac {pq}{2}}.} Conversely, any convex quadrilateral where the area can be calculated with this formula must be orthodiagonal. [ 5 ]
A perfect parallelepiped is a parallelepiped with integer-length edges, face diagonals, and space diagonals. In 2009, dozens of perfect parallelepipeds were shown to exist, [3] answering an open question of Richard Guy. One example has edges 271, 106, and 103, minor face diagonals 101, 266, and 255, major face diagonals 183, 312, and 323, and ...
The area of the parallelogram is the area of the blue region, which is the interior of the parallelogram. The base × height area formula can also be derived using the figure to the right. The area K of the parallelogram to the right (the blue area) is the total area of the rectangle less the area of the two orange triangles. The area of the ...
Quadrilaterals that are both orthodiagonal and equidiagonal, and in which the diagonals are at least as long as all of the quadrilateral's sides, have the maximum area for their diameter among all quadrilaterals, solving the n = 4 case of the biggest little polygon problem. The square is one such quadrilateral, but there are infinitely many others.
giving the basic form of Brahmagupta's formula. It follows from the latter equation that the area of a cyclic quadrilateral is the maximum possible area for any quadrilateral with the given side lengths. A related formula, which was proved by Coolidge, also gives the area of a general convex